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Chapter 9 - 9.8(a K PNO2 2 PNO PO2 2(b K PSbCl3 PCl2...

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9.8 (a) 2 2 2 NO 2 NO O = P K P P ; (b) 3 2 5 SbCl Cl SbCl = P P K P ; (c) 2 4 2 2 N H 2 N H = P K P P 9.20 r f 2 1 (CO , g) 394.36 kJ mol - ° = ∆ ° = - G G 3 1 1 1 69 ln ln 394.36 10 J mol ln 159.17 (8.314 J K mol )(298 K) 1.3 10 - - - ° = - ° = - - × = - = + = × G RT K G K RT K K In practice, no K will be so precise. A better estimate would be 69 1 10 × . Because Q < K , the reaction will tend to proceed to produce products. 9.32 H 2 (g) + Cl 2 (g) 2 HCl(g) 8 5.1 10 = × C K 2 2 2 [HCl] [H ][Cl ] = C K 3 2 8 3 2 3 2 2 8 3 12 1 2 (1.45 10 ) 5.1 10 [H ](2.45 10 ) (1.45 10 ) [H ] (5.1 10 )(2.45 10 ) [H ] 1.7 10 mol L - - - - - - × × = × × = × × = × 9.36 (a) 2 3 3 2 2 [NH ] 0.278 [N ][H ] = = C K 2 3 [0.122] 0.248 [0.417][0.524] = = C Q (b) C C Q K ; therefore the system is not at equilibrium. (c) Because < C C Q K , more products will be formed.
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9.58 Note: The volume of the system is not used because we are given pressures and K . Pressures (bar)N 2 (g) + 3 H 2 (g) 2 NH 3 (g) initial 0.025 0.015 0 change - x 3 - x 2 + x final 0.025 - x 0.015 3 - x 2 + x 3 2 2 2 2 NH 3 3 N H (2 ) 0.036 (0.025 )(0.015 3 ) = = = - - P x K P P x x Solving this explicitly will lead to a high-order equation, so first check to see if the assumption that 3 0.015 << x
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